# How do I graph the hyperbola with the equation 4x^2−y^2+4y−20=0??

Aug 14, 2018

Please see the explanation below

#### Explanation:

You need the center, the vertices and the asymptotes.

The equation is

$4 {x}^{2} - {y}^{2} + 4 y - 20 = 0$

Completing the square

$4 {x}^{2} - \left({y}^{2} - 4 y + 4\right) = 20 - 4 = 16$

Dividing by $16$

$\frac{4}{16} {x}^{2} - {\left(y - 2\right)}^{2} / 16 = 1$

${\left(x\right)}^{2} / 4 - {\left(y - 2\right)}^{2} / 16 = 1$

Comparing this to the general equation of a hyperbola

${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$

The center is $C = \left(h , k\right) = \left(0 , 2\right)$

The vertices are

$A = \left(h + a , k\right) = \left(0 + 2 , 2\right) = \left(2 , 2\right)$

and

$A ' = \left(h - a , k\right) = \left(0 - 2 , 2\right) = \left(- 2 , 2\right)$

And the asymptotes are

${\left(y - 2\right)}^{2} / 16 = {\left(x\right)}^{2} / 4$

$y - 2 = \pm \left(2 x\right)$

$y - 2 = 2 x$ and $y - 2 = - 2 x$

Now, you can plot the graph

graph{(4x^2-y^2+4y-20)(y-2-2x)(y-2+2x)=0 [-15.77, 16.27, -5.8, 10.2]}